Further results on common zeros of the solutions of two differential equations
© Asiri; licensee Springer 2012
Received: 8 May 2012
Accepted: 17 September 2012
Published: 4 October 2012
Two problems are discussed in this paper. In the first problem, we consider one homogeneous and one non-homogeneous differential equations and study when the solutions of these differential equations can have (nearly) the same zeros. In the second problem, we consider two linear second-order differential equations and investigate when the solutions of these differential equations can take the value 0 and a non-zero value at (nearly) the same points.
We apply the Nevanlinna theory and properties of entire solutions of linear differential equations.
In the first problem, the results determine all pairs of such equations having solutions with the same zeros or nearly the same zeros. Regarding the second problem, the results also show all pairs of such equations having solutions taking the value 0 and a non-zero value at (nearly) the same points.
There has been much research [1–8] on zeros of solutions of linear differential equations with entire coefficients. The principal paper  that was published in 1982 by Bank and Laine has stimulated many studies on this kind of problems. The reader is referred to [10–12] for background on some applications of the Nevanlinna theory. We use the standard notation of the Nevanlinna theory from .
In 1955, Wittich  proved the following theorem.
Theorem 1.1 If f is a non-trivial solution of , i.e., and is entire, then we have:
(ii) If f has finite order, then A is a polynomial.
Our previous paper  studied homogeneous linear differential equations having solutions with nearly the same zeros and proved several results, including the following.
Theorem 1.3 
Then is a constant and .
Recently, the same problem, but with non-homogeneous first-order differential equations, has been studied in , including the following result.
Theorem 1.4 
Then the following conclusions hold.
(I) If L is a rational function, then , L is a constant and .
(II) If L is a transcendental function, then one of the following cases holds: If, in addition, L has finite order in case (ii), then A, B, C, D are polynomials and so is .
(i) and v, w have no zeros.
where , and .
In this paper, our first result (Theorem 2.1) looks at the same problem but with one homogeneous and one non-homogeneous differential equations. In particular, we consider the first equation to be homogeneous of the second-order with a polynomial coefficient and the second equation to be non-homogeneous of the first-order with entire coefficients.
A further result (Theorem 2.2) studies the case where the solutions of two second-order homogeneous differential equations can take the value 0 and a non-zero value at (nearly) the same points.
2 Our results
Our first result is the following theorem.
Suppose that has finitely many zeros and poles (i.e., w and v have the same zeros with finitely many exceptions).
where and , are constants.
So, we have .
For example, let .
We now state our second result.
Theorem 2.2 Suppose that is a polynomial of degree n, and A is an entire function, and suppose that w solves (1) and v solves (3), and . Let and w have, with finitely many exceptions, the same zeros and the same multiplicities. Then one of the following holds.
(A) w has finitely many zeros and v is a polynomial and .
where σ, , , are non-zero constants.
Hence, has the same zeros as w. Here and .
Then , where .
Also , where .
Therefore, w and have the same zeros but the zeros are simple for w, double for . Here, and .
3 Proof of Theorem 2.1
The next step is to estimate the growth of M.
Claim 3.1 We claim that .
To show this, we know that since M has finitely many poles.
We also have .
This completes the proof of Claim 3.1.
because otherwise we can write to get a contradiction.
where , are polynomials.
where is rational.
where is rational and .
Thus, , H is entire, and B has no zeros.
Therefore, A is a polynomial.
where Q is a polynomial.
where is a constant and Q is a polynomial.
where is a constant and .
Now, from (23) and (24), we notice that w and v have the same zeros.
This completes the proof of Theorem 2.1. □
4 A lemma needed to prove Theorem 2.2
In order to prove Theorem 2.2, we must state and prove the following lemma. We include a proof for completeness.
Then there exists such that and , and for .
Proof The proof is by induction. It is obvious that the lemma is true when .
Now, we have two cases to consider.
Since we assumed the lemma is true for , there exists such that . But this contradicts our assumption that are distinct.
But this contradicts the fact that .
and and . □
5 Proof of Theorem 2.2
In particular, if w has finitely many zeros, then v is a polynomial, which gives . This completes the proof of part (A) in the conclusion.
Assume henceforth that w has infinitely many zeros. Then (26) implies that , and so A is a polynomial of degree at most n by the Wiman-Valiron theory . Also, since has infinitely many zeros.
Now, two cases have to be considered.
where , and ().
where δ is constant.
and , , β, −β, 0 cannot all be different.
We must now try six cases:
I(a): If and , then and . But this contradicts (30). Thus, this case cannot happen.
where , are constants, which yields . Putting this in (27) gives (7) with .
There are four more cases:
I(c): and .
I(d): and .
I(e): and .
I(f): and .
It is easy to check that case I(c) is impossible and cases I(d), I(e), I(f) all lead to (7) with .
From these cases, we find that , and so is non-constant. Also, we have (7) and case (B) of the conclusion.
where L is a rational function and Q is an entire function.
From (26), we have , and so Q is a polynomial.
Now, we have two cases to consider.
is a rational function, which is a contradiction since w has infinitely many zeros.
where is rational because is rational and is rational, and so is rational.
are rational functions and Q is a polynomial.
Now, we have two cases to consider:
which is a contradiction since w has infinitely many zeros.
where is a rational function.
So, G solves (1), and since and is a polynomial of degree n, we see that G is a transcendental entire function with finitely many zeros and has order .
where is a rational function.
Now, we can assume that because if , we can multiply w by .
is a rational function.
where and .
Claim 5.1 We claim that .
To show this, let .
where a is a constant. But this contradicts the fact that H and K are rational functions and G is a transcendental function. This completes the proof of Claim 5.1.
where , are polynomials, , and is non-constant because v is transcendental.
where , are rational functions and , are polynomials.
Therefore, because otherwise or . Thus, , both solve and have finitely many zeros, and they are linearly independent.
Hence, P is constant by , which contradicts our assumption in Case (II) that P is non-constant. □
The author would like to thank his supervisor Prof. Jim Langley for his support and guidance. Also, he would like to thank King Abdulaziz University for financial support for his PhD study.
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